Law of conservation of energy problems

[FarazAli]

The problem:
A mass m is attached to the end of a spring (constant k). The mass is given an initial displacement Xo from equilibrium, and an initial speed Vi. Ignoring friction and the mass of the spring, use energy methods to find (a) its maximum speed, and (b) its maximum stretch from equilibrium, in terms of the given quantities

The answer I got for part a:
$$v_{f} = \left(\frac{(kx_{0}^2 + mv_{i}^2)}{m}\right)^{1/2} = \left(\frac{kx_{o}^2}{m}\right)^{1/2} + v_{i}$$

- I set $$PE_{i} + KE_{i} = PE_{f} + KE_{f}$$ and crossed out $$PE_{f}$$ because max velocity occurs at zero potential

The answer I got for part b:
$$x = \left({\frac{mv_{i}^2}{k}}\right)^{1/2} = v_{i}\left(\frac{m}{k}\right)^{1/2}$$

- Again I used law of conservation of energy equation, and crossed $$KE_{f}$$ and $$PE_{i}$$ because the max stretch occurrs when $$PE_{f}$$ is maximum, therefore $$KE_{f} = 0$$. The box's $$PE_{i} = 0$$ ( starting off from part a )

What I want to know is whether these answers are correct, as I have absolutely no other way, except for my own intillect, to find out.

 

[Doc Al]

The answer I got for part a:
$$v_{f} = \left(\frac{(kx_{0}^2 + mv_{i}^2)}{m}\right)^{1/2} = \left(\frac{kx_{o}^2}{m}\right)^{1/2} + v_{i}$$

This part is correct:
$$v_{f} = \left(\frac{(kx_{0}^2 + mv_{i}^2)}{m}\right)^{1/2}$$
But that does not equal
$$\left(\frac{kx_{o}^2}{m}\right)^{1/2} + v_{i}$$
Realize that $(a^2 + b^2)^{1/2} \ne a + b$.

- I set $$PE_{i} + KE_{i} = PE_{f} + KE_{f}$$ and crossed out $$PE_{f}$$ because max velocity occurs at zero potential

The method is correct.

The answer I got for part b:
$$x = \left({\frac{mv_{i}^2}{k}}\right)^{1/2} = v_{i}\left(\frac{m}{k}\right)^{1/2}$$

This is incorrect. Your answer for b should be very similar to the answer for a, except that you solve for maximum X, instead of maximum V.

 

[FarazAli]

silly mistake. Thanks.

So I redid the second part and got this:
$$\left(\frac{kx_{0}^2 + mv_{i}^2}{k}\right)^{1/2}$$

 

[Doc Al]

Looks good to me.

 

[FarazAli]

thanks for the help